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The probability that there is at least one error is an accounts statement prepared by A is 0.2 and for B and C they are 0.25 and 0.4 respectively A, B, and C prepared 10, 16 and 20 statements respectively. Find the expected number of correct statements in all
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Solution | |
Let "x", "y" and "z" represent the number of correct statements prepared by "A", "B" and "C" respectively. For "A" In the experiment of testing whether a statement prepared by "A" is correct or not, there are two possible events,
The probability that there is at least one error is an accounts statement prepared by A is 0.2
Considering the two events of "Error" or "No Error" to be the only possibilities, they are exhaustive events
Since there can either be atleast one error or there cannot be any error, the two events of "Error" or "No Error" are mutually exclusive
From (1) and (2) we can write
No. of repetitions of the experiment = No. of statements made by "A" tested
Expected no. of correct statements prepared by "A" = Expected frequency of occurance of the event "No Error" in "N" repetitions of
⇒ E(x) = 8 Similarly For "B" Probability that a statement prepared by "B"
No. of repetitions of the experiment = No. of statements made by "B" tested
Expected no. of correct statements prepared by "B" = Expected frequency of occurance of the event "No Error" in "N" repetitions of
⇒ E(y) = 12 Similarly For "C" Probability that a statement prepared by "C"
No. of repetitions of the experiment = No. of statements made by "C" tested
Expected no. of correct statements prepared by "C" = Expected frequency of occurance of the event "No Error" in "N" repetitions of
⇒ E(z) = 12 Expected no. of correct statements in all prepared by "A", "B" and "C" together
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Credit : Vijayalakshmi Desu |